A bar magnet with poles `sqrt3cm` apart and pole strength `14*4Am` rests with its centre on a frictionless pivot. The magnet is held in equilibrium at an angle of `60^@` to a uniform magnetic field of intensity `0*20T` by applying a force F at right angles to its axis, `0*1m` from its pivot. What is the value of force F?

Answer

`0*432N`

Explanation

Here, `2l=sqrt3cm=sqrt3xx10^-2m`
`m=14.4Am`, `theta=60^@`, `B=0*20T`
`F=?` `r=0*1m`
Now, `tau=rFsin90^@=MBsintheta`
`F=(MBsintheta)/(r)=(m(2l)Bsintheta)/(r)`
`F=(14.4xxsqrt3xx10^-2xx0*2sin60^@)/(0*1)`
`=0*432N`